递归斐波那契汇编 [英] Recursive fibonacci Assembly

问题描述

今天，我在汇编中编写了递归斐波那契，但它不起作用.我使用NASM将其编译为目标文件，然后使用gcc将其制成精灵.
当我输入1或2时，该功能可以正常运行，但是当我输入3、4、5、6或更多时，该功能不起作用.我认为函数调用自身存在问题.

Today I wrote a recursive fibonacci in assembly and it doesn't work. I compiled it to object file with NASM and than made it elf with gcc.
When I enter 1 or 2 the function works perfectly, but when I enter 3, 4, 5, 6, or more the function doesn't work. I think there is problem where the function calls itself.

此代码:

SECTION .data ;init data




str: db "This equal: %d",10,0

SECTION .text   ;asm code


extern printf
global main

main:
push ebp
mov ebp,esp
;--------------------


push 03  ; the index 
call _fibonacci
add esp,4

push DWORD eax
push str
call printf


;---------------------

mov esp,ebp
pop ebp
ret

这是功能:

_fibonacci:

push ebp
mov ebp,esp


mov ebx, [ebp+8] ;; param n 
cmp ebx,0
jne CHECK2

mov eax,0
jmp _endFIbofunc        

CHECK2: 
    cmp ebx,0x1
    jne ELSE3
    mov eax,1
jmp _endFIbofunc

ELSE3:

mov ebx,[ebp+8] 
dec ebx  ;; n-1


;;  FIRST call
push ebx
call _fibonacci
add esp,4
mov edx,eax

;;  SEC CALL
dec ebx
push ebx
call _fibonacci
add esp,4 
add eax,edx


mov eax,[ebp-4]

_endFIbofunc:

mov esp,ebp
pop ebp
ret

在Ubuntu 16.04上运行它后，它发送错误:

After I ran it on Ubuntu 16.04 it send error:

分段错误(核心已转储)

Segmentation fault (core dumped)

可能是什么问题?

推荐答案

除了提供的其他答案之外，还有另一种解决方法:

In addition to the other answers provided, here's an alternate solution:

_fibonacci:
        mov     eax,[esp+4]             ;eax = n
        cmp     eax,2                   ;br if n < 2
        jb      _endFIbofunc
        dec     eax                     ;push n-1
        push    eax
        call    _fibonacci              ;returns eax = fib(n-1)
        xchg    eax,[esp]               ;eax = n-1, [esp] = fib(n-1)
        dec     eax                     ;push n-2
        push    eax
        call    _fibonacci              ;returns eax = fib(n-2)
        add     eax,[esp+4]             ;eax = fib(n-1)+fib(n-2)
        add     esp,8
_endFIbofunc:
        ret

琐事-fib(47)是最大的<2 ^ 32.递归调用的数量= 2 * fib(n + 1)-1.

Trivia - fib(47) is the largest < 2^32. The number of recursive calls = 2*fib(n+1)-1.

 n     fib(n)      # calls

 0          0            1
 1          1            1
 2          1            3
 3          2            5
 4          3            9
 5          5           15
 6          8           25
 7         13           41
 8         21           67
 9         34          109
10         55          177
11         89          287
12        144          465
13        233          753
14        377         1219
15        610         1973
16        987         3193
17       1597         5167
18       2584         8361
19       4181        13529
20       6765        21891
21      10946        35421
22      17711        57313
23      28657        92735
24      46368       150049
25      75025       242785
26     121393       392835
27     196418       635621
28     317811      1028457
29     514229      1664079
30     832040      2692537
31    1346269      4356617
32    2178309      7049155
33    3524578     11405773
34    5702887     18454929
35    9227465     29860703
36   14930352     48315633
37   24157817     78176337
38   39088169    126491971
39   63245986    204668309
40  102334155    331160281
41  165580141    535828591
42  267914296    866988873
43  433494437   1402817465
44  701408733   2269806339
45 1134903170   3672623805
46 1836311903   5942430145
47 2971215073   9615053951
48 4807526976   ...

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